requesting for explanation of surface integration

Jan 2017
209
3
Toronto
Evaluate ∫∫ < x, y, -2 > * N dS, where D is given by z = 1 - x^2 - y^2, x^2 + y^2 <= 1, oriented up.

Official Answer: - pi

Cross Product:
\(\displaystyle
\int_{0}^{2\pi} \int_{0}^{1} ( r cos \theta , r sin \theta , -2 ) * ( 2r^2 cos \theta , 2r^2 sin \theta , r ) ~ dr d \theta
\)

\(\displaystyle
\int_{0}^{2\pi} \int_{0}^{1} ( 2r^3 - 2r ) ~dr d \theta = -\pi
\)

I managed to get the correct answer. but I didn't understand why the Jacobian of 'r' was missing from the integration since I used polar coordinate.
 

romsek

Math Team
Sep 2015
2,885
1,609
USA
$n dS = (2x,2y,1) \to (2 r \cos(\theta), 2r \sin(\theta), 1)$

for some reason you added a factor of $r$ in which accounts for the missing $r$ in your integral.
 
Jan 2017
209
3
Toronto
$n dS = (2x,2y,1) \to (2 r \cos(\theta), 2r \sin(\theta), 1)$

for some reason you added a factor of $r$ in which accounts for the missing $r$ in your integral.
\(\displaystyle
G(r, \theta) = < r cos \theta , r sin \theta , 1 - r^2 >
\)
\(\displaystyle
dG/dr = < cos \theta , sin \theta , -2r >
\)
\(\displaystyle
dG/d \theta = < - r sin \theta , r cos \theta , 0 >
\)
Cross Product
\(\displaystyle
dG/dr ~X~ dG/d \theta = < 2r^2 cos \theta , 2r^2 sin \theta , r >
\)

Dot Product
\(\displaystyle
\int_{0}^{2\pi} \int_{0}^{1} < r cos \theta , r sin \theta, -2 > * < 2r^2 cos \theta , 2r^2 sin \theta , r >
\)
 

romsek

Math Team
Sep 2015
2,885
1,609
USA
\(\displaystyle
G(r, \theta) = < r cos \theta , r sin \theta , 1 - r^2 >
\)
\(\displaystyle
dG/dr = < cos \theta , sin \theta , -2r >
\)
\(\displaystyle
dG/d \theta = < - r sin \theta , r cos \theta , 0 >
\)
Cross Product
\(\displaystyle
dG/dr ~X~ dG/d \theta = < 2r^2 cos \theta , 2r^2 sin \theta , r >
\)
I really don't think you can start in polar coordinates like that. Not using the standard definition of the surface integral. There is an equivalent definition but the formula is different.

You have to find the normal vector in cartesian coordinates first, dot it with the field vector, then you can convert everything over to polar coordinates for the integration.
 
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